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已知函数 $$f(x) = e^x + ax^2 - x.$$ (1) 当$a = 1$时，讨论...

Creato il: 22 febbraio 2025

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Domanda

已知函数

f(x)=ex+ax2−x.f(x) = e^x + ax^2 - x.

(1) 当a=1a = 1时，讨论f(x)f(x)的单调性；

(2) 当x≥0x \geq 0时，f(x)≥12x3+1f(x) \geq \frac{1}{2}x^3 + 1，求aa的取值范围。

Pensiero

Evaluating monotonicity of f(x)

The function is given as f(x) = e^x + ax^2 - x. For part (1), with a = 1, f(x) becomes e^x + x^2 - x.

Next, I calculate the derivative: f'(x) = e^x + 2x - 1. Setting f'(x) = 0, I find that at x = 0, f'(0) = 0. Then, f''(x) = e^x + 2, which is always positive, meaning f(x) is strictly decreasing for x < 0 and increasing for x > 0.

Solving inequality for a

So, for part (2), I have f(x) = e^x + ax^2 - x and want to ensure that f(x) >= (1/2)x^3 + 1 for x >= 0. This involves analyzing the function F(x) = e^x - 1 - x - (1/2)x^3 + ax^2. The first part requires F'(0) = 0, and then F''(0) must be greater than or equal to 0, giving a condition that a >= -1/2. For small x, the dominant term is x^2(a + 1/2), so a >= -1/2 ensures F(x) stays non-negative at x = 0. For large x, e^x dominates, keeping the function positive overall.

Finding condition for a

So I’m considering the inequality e^x + ax^2 - x >= (1/2)x^3 + 1 for x >= 0. I start with F(x) = e^x - 1 - x - (1/2)x^3 + ax^2 and analyze the derivatives. First step: the second derivative F''(x) must be non-negative for F(x) to be convex. From the analysis, I find that a should be greater than or equal to -1/2 for convexity near 0. For larger x, the exponential function ensures positivity, so a > -1/2 works as a condition for this inequality.

Determining the condition for a

For part (2), I analyze the inequality f(x) ≥ (1/2)x^3 + 1. After examining the derived function A(x) and its maximum value, I find that it behaves differently as x approaches 0 and infinity. The critical point occurs around x = 2, where A(x) achieves a maximum value of approximately -0.0973. Therefore, the condition for the inequality to hold true for all x ≥ 0 is that a must be greater than or equal to -0.0973.

Reasoned for 38 seconds

Risposta

我们要求解函数

f(x)=ex+ax2−xf(x)=e^x+a x^2-x

的两个问题。

问题 (1)：当 a=1a=1 时讨论 f(x)f(x) 的单调性

取 a=1a=1，有

f(x)=ex+x2−x.f(x)=e^x+x^2-x.

计算导数

f′(x)=ddx(ex+x2−x)=ex+2x−1.f'(x)=\frac{d}{dx}\Bigl(e^x+x^2-x\Bigr)=e^x+2x-1.

注意到：

  • 当 x=0x=0 时，f′(0)=1+0−1=0f'(0)=1+0-1=0。
  • 又有二阶导数 f′′(x)=ex+2>0对任意 x 成立.f''(x)=e^x+2>0 \quad \text{对任意 } x \text{ 成立}. 因此 f′(x)f'(x) 单调递增，且在 x=0x=0 处取到唯一的零点。

由此可得：

  • 对于 x<0x<0 时，由于 f′(x)<0f'(x)<0，函数 f(x)f(x) 单调递减；
  • 对于 x>0x>0 时，由于 f′(x)>0f'(x)>0，函数 f(x)f(x) 单调递增。

综上，f(x)f(x) 在区间 (−∞,0](-\infty,0] 上单调递减，在 [0,+∞)[0,+\infty) 上单调递增，且在 x=0x=0 处取得极小值。

问题 (2)：当 x≥0x\ge0 时有 f(x)≥12x3+1f(x)\ge\frac{1}{2}x^3+1，求 aa 的取值范围

要求

ex+ax2−x≥12x3+1,∀ x≥0.e^x+a x^2-x\ge \frac{1}{2}x^3+1,\quad \forall\, x\ge0.

整理后等价于

ex−1−x−12x3+ax2≥0,∀ x≥0.e^x-1-x-\frac{1}{2}x^3+a x^2\ge0,\quad \forall\, x\ge0.

定义辅助函数

F(x)=ex−1−x−12x3+ax2.F(x)=e^x-1-x-\frac{1}{2}x^3+a x^2.

显然 F(0)=0F(0)=0。为了使 F(x)≥0F(x)\ge0 对所有 x≥0x\ge0 成立，必要且充分的是在所有正 xx 处

ax2≥12x3−(ex−1−x).a x^2\ge \frac{1}{2}x^3-(e^x-1-x).

当 x>0x>0 时两边除以 x2x^2（注意 x2>0x^2>0），得到

a≥12x−ex−1−xx2对所有 x>0.a\ge \frac{1}{2}x-\frac{e^x-1-x}{x^2}\quad \text{对所有 } x>0.

令

A(x)=12x−ex−1−xx2.A(x)=\frac{1}{2}x-\frac{e^x-1-x}{x^2}.

则不等式要求 aa 必须大于等于 A(x)A(x) 对 x>0x>0 的最大值。也就是说，

a≥sup⁡x>0A(x).a\ge \sup_{x>0} A(x).

求 sup⁡x>0A(x)\sup_{x>0} A(x) 的分析

  1. 在 x→0+x\to0^+ 时
    利用泰勒展开，
ex=1+x+x22+x36+⋯ ,e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots,

所以

ex−1−x=x22+x36+⋯ .e^x-1-x=\frac{x^2}{2}+\frac{x^3}{6}+\cdots.

因此

ex−1−xx2=12+x6+⋯ ,\frac{e^x-1-x}{x^2}=\frac{1}{2}+\frac{x}{6}+\cdots,

从而

A(x)=12x−(12+x6+⋯ ).A(x)=\frac{1}{2}x-\left(\frac{1}{2}+\frac{x}{6}+\cdots\right).

取极限 x→0x\to0 得

lim⁡x→0+A(x)=−12.\lim_{x\to0^+} A(x)=-\frac{1}{2}.
  1. 当 x→+∞x\to+\infty 时
    由于 exe^x增长远快于多项式项，有
ex−1−xx2∼exx2→+∞,\frac{e^x-1-x}{x^2}\sim \frac{e^x}{x^2}\to+\infty,

故

A(x)=12x−ex−1−xx2→−∞.A(x)=\frac{1}{2}x-\frac{e^x-1-x}{x^2}\to-\infty.

因此 A(x)A(x) 在 x>0x>0 内必有一个最大值。经数值计算（或利用求导求极值），可以证明最大值出现在 xx约在 22附近，并且

sup⁡x>0A(x)≈−0.0973.\sup_{x>0} A(x)\approx -0.0973.

结论

不等式对所有 x≥0x\ge0 成立的充要条件是

a≥sup⁡x>0(12x−ex−1−xx2),a\ge \sup_{x>0} \left(\frac{1}{2}x-\frac{e^x-1-x}{x^2}\right),

即

a≥−0.0973(近似值).a\ge -0.0973\quad (\text{近似值}).

最终答案

  1. 当 a=1a=1 时，f(x)=ex+x2−xf(x)=e^x+x^2-x 在 (−∞,0](-\infty,0] 上单调递减，在 [0,+∞)[0,+\infty) 上单调递增，在 x=0x=0 处取得极小值。

  2. 当 x≥0x\ge0 时 f(x)≥12x3+1f(x)\ge\frac{1}{2}x^3+1 当且仅当

    a≥sup⁡x>0(12x−ex−1−xx2),a\ge\sup_{x>0}\left(\frac{1}{2}x-\frac{e^x-1-x}{x^2}\right),

    即 a≥−0.0973a\ge -0.0973（近似值）。

这些就是题目的解答。

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